Question: Solve for $x$ : $3x^2 + 36x + 108 = 0$
Dividing both sides by $3$ gives: $ x^2 + {12}x + {36} = 0 $ The coefficient on the $x$ term is $12$ and the constant term is $36$ , so we need to find two numbers that add up to $12$ and multiply to $36$ The number $6$ used twice satisfies both conditions: $ {6} + {6} = {12} $ $ {6} \times {6} = {36} $ So $(x + {6})^2 = 0$ $x + 6 = 0$ Thus, $x = -6$ is the solution.